Question

Which of the following is not the value of $2\,\tan^{-1}x$ ?

• A. $\tan^{-1}\frac{2x}{1+x^2}$
• B. $\tan^{-1}\frac{2x}{1-x^2}$
• C. $\cos^{-1}\frac{1-x^2}{1+x^2}$
• D. $\sin^{-1}\frac{2x}{1+x^2}$

Answer 1

Answer: (A)

Since $ \frac{2\,tan\,\theta}{1-tan^{2}\theta} = tan \,2\,\theta$
$\frac{ 1-tan ^{2}\theta}{1+tan^{2}\theta} = cos \,2\,\theta,\frac{ 2\,tan \,\theta}{1+tan\,\theta} = sin \,2\,\theta$
$ \therefore \left(b\right)= 2\,\theta, \left(c\right)=2\,\theta, \left(d\right)= 2\,\theta = 2\,tan^{-1}x $
Hence $tan^{-1} \frac{2x}{1+x^{2}} \ne 2\,tan^{-1} x$